For matrix A, A^-1 = kA. What is k?

Determinants, Adjoint & Inverse · Question 67 of 87

Question 67

HardOriginal question

For $A = \begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}$ , $A^{-1} = kA$ . What is $k$ ?

- \frac{18}{19}

Correct answerYour answer

- \frac{1}{19}

Correct answerYour answer

\frac{1}{38}

Correct answerYour answer

\frac{1}{19}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$|A| = -4 - 15 = -19$ and $\operatorname{adj} A = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = -A$ .
$A^{-1} = \frac{1}{-19}(-A) = \frac{1}{19}A$ , so $k = \frac{1}{19}$ . (Indeed $A^2 = 19I$ .)

Worth noting

$-\frac{1}{19}$ divides by $|A|$ but forgets that the adjoint is $-A$ , not $A$ .

Ten questions from Determinants, Adjoint & Inverse, at your own pace.

Ten questions from Determinants, Adjoint & Inverse, at your own pace.